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Course: MATHEMATIC 55, Summer 2009
School: Berkeley
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Word Count: 507

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55 Math Worksheet Adapted from worksheets by Rob Bayer, Summer 2009. Sets, Functions, Cardinality 1. Determine whether each of the following are injective, surjective, both (bijective), or neither. Prove your answer (a) f : N N f (x) = x2 Injective (b) f : R R f (x) = 3x + 4 Bijective (c) f : (0, 1) (1, ) f (x) = 1 x Biective 2. For a function f : A B, given S A, let f (S) := {y B : x S, f (x) = y}. Prove...

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55 Math Worksheet Adapted from worksheets by Rob Bayer, Summer 2009. Sets, Functions, Cardinality 1. Determine whether each of the following are injective, surjective, both (bijective), or neither. Prove your answer (a) f : N N f (x) = x2 Injective (b) f : R R f (x) = 3x + 4 Bijective (c) f : (0, 1) (1, ) f (x) = 1 x Biective 2. For a function f : A B, given S A, let f (S) := {y B : x S, f (x) = y}. Prove that if f is one-to-one, then f (S T ) = f (S) f (T ) for any S, T A. We have to show that the two SETS f (S T ) and f (S) f (T ) are equal. So we have two steps: 1) We show f (S T ) f (S) f (T ): Let b f (S T ). Then this means that there exists an x S T such that b = f (x). Thus x S and x T . So f (x) f (S) and f (x) f (T ) Therefore f (x) f (T ) f (S). But b = f (x), so therefore b f (S) f (T ) which is what we wanted to show. 2) We show next that f (S) f (T ) f (S T ): Let b f (S) f (T ). Then b f (S) and b f (T ). Thus there exists x1 S and x2 T such that f (x1 ) = b and f (x2 ) = b. (Think about this step!) But then since f is injective, x1 = x2 , and they are both in S T . Therefore, b = f (x1 ) f (S T ). Then, give an example of a non-injective function f , and sets S, T which for this is not true. What's our favorite example of a non-injective function? f (x) = x2 . Try picking S = [-2, 1] and T = [-1, 2] be intervals (subsets of R.) Then S T = [-1, 1] and we have : f (S T ) = [0, 1] (Make sure you see why this is true) But f (S) = [0, 4] and f (T ) = [0, 4], so f (S) f (T ) = [0, 4]. 3. Use the definitions of the set complement, union and intersection to prove that A B = A B 4. Decide whether each of the following are true or false. For those that are true, prove it. For those that are false, provide a counterexample.See Solutions 4 (a) If f and g are injective, then so is f g. (b) If f and g are surjective, then so is f g. (c) If g f is injective, then so is f . (d) If g f is injective, then so is g. (e) If f is not surjective, then g f is not surjective. 5. Even when a function f : A B is not invertible, we will often talk about the inverse image of a set C B, which we denote by f -1 (C) and which is defined as {x|f (x) C}. Determine each of the following: See Solutions 4 (a) f -1 ({September12}) where f :(People) (days of year), f (x) = x's birthday. (b) f -1 ({x|x > 3}) where f : R R; f (x) = x2 . (c) f -1 ({3}) where f : P(N) N, f (S) = min(S) (or 0 if S = ).
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